Finite index subgroup of $\mathbb{Z}^2$ that is invariant under a non-singular matrix

Question

Let $M $ be a matrix in $ \operatorname{GL}(2, \mathbb{Z})$ that has at least one eigenvalue of absolute value strictly bigger than $1$. What are the finite index subgroups $H$ of $\mathbb{Z}^2$ such that for all $v \in H$, we have $M(v) \in H$?

Question: One example of such subgroup has the form $H=\langle (a,0),(0,a)\rangle$, for a non-zero integer $a$, this example works for any matrix in $ \operatorname{GL}(2, \mathbb{Z})$. I was wondering if there are any other such subgroups.

Thought so far: I think the finite index subgroups of $\mathbb{Z}^2$ have the form $H = \langle (a,b),(c,d)\rangle$, where $a,b,c,d \in \mathbb{Z}$ such that $ad-bc\neq 0$. Hence, in order for $H$ to be a subgroup, we need $M(a,b)= \alpha_1(a,b) + \alpha_2(c,d)$ and $M(c,d)= \beta_1(a,b) + \beta_2(c,d)$ for some integers $\alpha_i$ and $\beta_i$, but I can't see any obvious solutions for these equations.

I was also thinking maybe we can use the eigenvectors, but unfortunately the eigenvectors might not have integers entries.

Any ideas for constructing such a subgroup would be really appreciated.

Answer 1

If $v$ is not scalar, the ring $\mathbb Z[M]$ generated by $M$ has rank $2$ over $\mathbb Z$. It is therefore either an order in a number field, a finite-index subring of $\mathbb Z^2$, or $\mathbb Z[\epsilon]/\epsilon^2$.

$\mathbb Z^2$ is then a module over $\mathbb Z[M]$, and you are looking to describe the finite-index submodules of $\mathbb Z[M]$.

In the simplest case, $\mathbb Z^2$ is a (locally) free module over $\mathbb Z[M]$, and so you are describing the ideals of $\mathbb Z[M]$. In the ring of integers in a number field case, this has a particularly simple description, as products of prime ideals.

In the general order case, submodules are given by products of prime ideals away from those primes where the order $\mathbb Z[M]$ is not maximal or the module $\mathbb Z^2$ is not locally free, and there is additional complexity at those primes. Said another way, the simplest construction of such a subgroup is to take the product of $\mathbb Z^2$ with some ideal of the order $\mathbb Z[M]$, for example the intersection of $\mathbb Z[M]$ with an ideal of the ring of integers of the number field.

In the other cases there is a similarly concrete description.

Edit: Actually maybe I should say this a different way. Your idea of using the eigenvectors is a good one, but it's better to use the eigenvectors modulo $p$. Consider the characteristic polynomial of the element $M$, and its discriminant. If this discriminant is nonzero then there are infinitely many primes $p$ modulo which the discriminant is a nonzero square (quadratic reciprocity + Dirichlet's theorem). When the entries of $M$ are taken mod $p$, allowing us to view $M$ as a matrix over $\mathbb F_p$, it has two distinct eigenvalues and therefore two eigenvectors. The subgroup of vectors congruent mod $p$ to a multiple of one of these two eigenvectors is an index $p$ subgroup invariant under $M$.

If the discriminant is $0$ and $M$ is non-scalar, so $M$ has a single $2 \times 2$ Jordan block, then modulo all but finitely many primes $p$, $M$ will still have a single $2\times 2$ Jordan block, and you can take elements congruent mod $p$ to multiples of the unique eigenvector. Modulo the other primes, where $M$ becomes scalar, you can take multiples of any vector.

Answer 2

To illustrate how such a question is of arithmetic nature (and what a complete answer should look like), here is a partial answer in a specific case.

Namely: I specify to $M=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$ and I address the question: what are invariant finite index subgroups of prime index.

The characteristic polynomial of this matrix is $X^2-3X+1$, which has discriminant 5. By the quadratic reciprocity formula, for an odd prime $p\neq 5$, 5 is a square mod $p$ iff $p=\pm 1$ mod $5$ (i.e. the decimal expansion of of $p$ terminates with $1$ or $9$). And this polynomial is irreducible mod $2$, and has a double root mod $5$.

Hence, for a prime $p$:

• for $p=\pm 2$ mod $5$, there is no $M$-invariant subgroup of index $p$ in $\mathbf{Z}^2$;
• for $p=\pm 1$ mod $5$, there are exactly two $M$-invariant subgroups of index $p$ in $\mathbf{Z}^2$;
• there is a single $M$-invariant subgroup of index $5$ in $\mathbf{Z}^2$.

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Remarks:

For a general matrix $M$, the characteristic polynomial is $X^2+nX\pm 1$ for some $n$ and there should be a similar discussion.

For a general index $q$, one should boil down to when $q$ is a power of a prime $p$. Then in turn one should boil down to when the quotient is cyclic (the quotient being isomorphic to $C_{p^a}\times C_{p^b}$ for some $a\le b$, the subgroup is contained in $p^a\mathbf{Z}^2$ and we can then "replace" $\mathbf{Z}^2$ with $p^a\mathbf{Z}^2$ to assume $a=1$).

Answer 3

For A fixed $M$, such finite index subgroups may be categorized as follows. Let $W\in GL(2,\mathbb{Z})$ and $b$ a positive integer dividing the lower left element of the matrix $W^{-1} M W$. Then the columns of the matrix $W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}$ generate a subgroup $H$ of $\mathbb{Z}^2$ (of finite index $b$) for which $Mv\in H$ for any $v\in H$. Conversely, all such subgroups for this $M$ arise this way, up to scaling the subgroup by an integer $a$ as already noted in the question.

To see this, first note that a finite index subgroup of $\mathbb{Z}^2$ may be generated by the columns of a non-singular integer matrix $A$, known as a basis. Any other basis equals $AU$, where $U$ is unimodular, i.e. an element of $GL(2,\mathbb{Z})$.

For unimodular $M$, you want to know for which such $A$ do we have $MA = AV$, for some integer matrix $V$, as such $V$ just makes integer linear combinations of the columns of the basis $A$, which give elements of the finite index subgroup $H$. But determinants then imply $V$ is also unimodular, as it has the same determinant as $M$. So $M$ takes any basis of $H$ to another basis of $H$.

Integer matrices have a Smith Normal Form: there exist unimodular $W,V$ so that $W\begin{bmatrix} a & 0\\0 & ab\end{bmatrix}V = A$, where $a,b$ are integers (here positive, as the determinant is non-zero). As any choice of basis will do, we choose the basis $W\begin{bmatrix}a & 0\\0 & ab\end{bmatrix}$.

So an $H$ may be specified by a unimodular $W$ and integers $a$ and $b$. As $a$ simply scales everything we henceforth assume WLOG that $a=1$.

We now characterize such pairs $W, b$. We have from before the equation $$ M (W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}) = (W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}) V$$ so $$ (W^{-1} M W) \begin{bmatrix} 1 & 0\\0 & b\end{bmatrix} = \begin{bmatrix} 1 & 0\\0 & b\end{bmatrix} V$$

As $W$ is unimodular, it has integer inverse, so all matrices above are integer. The matrix on the left has second column divisible by $b$, while the matrix on the right has second row divisible by $b$. As they equate, both sides must equal a matrix of the form $\begin{bmatrix} x & by\\bz & bw\end{bmatrix}$ where $w,x,y$ and $z$ are integers. But then $V$ equals $\begin{bmatrix} x & by\\z & w\end{bmatrix}$, so $V$ unimodular implies $xw - byz = \pm 1$.

Similarly, we have $W^{-1} M W = \begin{bmatrix} x & y\\bz & w\end{bmatrix}$, since the diagonal matrix on the right simply scales the second column by $b$. So we started with a general subgroup $H$ invariant under $M$ - generated by the columns of a matrix $A$ - and showed that this $H$ has a basis of the form $W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}$ where the index $b$ divides the bottom left entry of $W^{-1} M W$, as claimed.

For the other direction, let $b$ be a positive factor of the bottom left entry of $W^{-1} M W$, where $W$ is any unimodular matrix. This means

$$W^{-1} M W = \begin{bmatrix} x & y\\bz & w\end{bmatrix}$$

for some integers $w, x, y$ and $z$, where $xw - bzy =\pm 1$, as the matrix is unimodular. But then $$ M W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix} = W \begin{bmatrix} x & y\\bz & w\end{bmatrix}\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix} $$ and $$M \left(W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}\right)= W\begin{bmatrix} x & by\\bz & bw\end{bmatrix}=W \begin{bmatrix} 1 & 0\\0 & b\end{bmatrix} \begin{bmatrix} x & by\\z & w\end{bmatrix}= \left(W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}\right) V $$

where the matrix $V$ is unimodular, as we have already noted $xw - bzy =\pm 1$. But this means the subgroup $H$ of $\mathbb{Z}^2$ generated by the columns of $ W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}$ is invariant under $M$, as claimed.